Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 3 (Q.No. 36)
36.
The ratio of molar specific heats for monoatomic gas is
Discussion:
12 comments Page 1 of 2.
Pratik said:
4 years ago
For mono atomic Cp/Cv=5/3 = 1.66.
For Di atomic Cp/Cv= 1.40 (Cp = 7/2 * R Cv = 5/2*R)
For tri atomic Cp/Cv=1.33 (Cp = 4*R Cv = 3 * R).
For Di atomic Cp/Cv= 1.40 (Cp = 7/2 * R Cv = 5/2*R)
For tri atomic Cp/Cv=1.33 (Cp = 4*R Cv = 3 * R).
Sudheer said:
6 years ago
The ratio γ = 1.66 for an ideal monoatomic gas and γ = 1.4 for air, which is predominantly a diatomic gas.
Srinu@v said:
7 years ago
The kinetic energy of one molecule in a particular direction.
Krupa said:
7 years ago
How to find DOF for gas?
Krupa said:
7 years ago
How to find DoF?
S Bera said:
8 years ago
X=4 dof for monoatomic, x=5 dof for diastolic, x=6 dof for triatomic.
RISHI said:
9 years ago
1 + 2/f where f is the deg of freedom and for monoatomic gas DOF IS 3 hence 1 + 2/f = 1.67. So it is the answer.
Rayudu said:
1 decade ago
1+2/x.
Where,
x = 3 for monotonic.
x = 5 for diatomic.
x = 6 for triatomic.
And x is degrees of freedom.
Where,
x = 3 for monotonic.
x = 5 for diatomic.
x = 6 for triatomic.
And x is degrees of freedom.
NAVIN said:
1 decade ago
Molar specific heat at constant pressure = 5/2*R.
Molar specific heat at constant volume is = 3/2*R
So Pm/Vm = 5/2*2/3 = 5/3 = 1.67
Molar specific heat at constant volume is = 3/2*R
So Pm/Vm = 5/2*2/3 = 5/3 = 1.67
Vaishali said:
1 decade ago
Tell me anyone the correct formula for finding the ratio of molar specific heat of gases.
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